III 1.(0,5 diem) T\ mm de phi^oTnjj tnnh co hai nghi^m
1.Xac din hm de ham so y= Vm" 4x (m^ 0) xac djnh tren nuTa khoang
(-2; 1|.
2 . T i m lap xac dinh ciia ham so y = '
Cau M. (2,0 diem)
1. Xiic dinli a dc ba diA-Int: lhang d|: y = 5(x + 1), d , ; y = - 3 x - ( a + 6) va
d^: y = ax + 2a + 3pliaii bicl va dung quị ^
2. Xac diiih ham so y = 2x~ + ax - b bicl do ihi ham só nay la parabol c6 dính I ( - h 4).
Cau m. (2,0 diem) *
I 1. Cho a, b, c la ba só thifc ihoa man 5a + 2h + 3c = (). ChiJng minh rang
phiftfng I n n h ax" + bx + c = 0 c6 nghicm.
2. Giiii phiAJng Irinh X" + X - 12 =: 2 x ( x - 3 ) . i
3. Giai phifdng Irinh (x" - 3 x ) \ / 4 - x -3x~ =0.
CAM \. (2,0 diem).
TrongmatphangOxycho A ( - l ; -1), B((); I ) , C(4; - i), D(3; - 3 ) ! 1. Chírng minh liV giac ABCD la hinh chiT nhiil.
2. Tinh clui \ va diC'n lich ciia hlnh chu" nhat dọ
Cau V. (2,0 diem)
'2x~.3y + /,=:5
x + y - 3/ = 7 (khong su" dung may tinh cam lay). 3x + y - 4/ = 2
1. Giiii he phiftJng tnnh
, 7x - 9
2. Tnn gia tri k'Jn nhal cíia y = vdi x >9. 2x
Cau VỊ (1,0 diem). Cho lam giac ABC c6 AB = 2, BC = 4, CA = 3. Tinh: 1. Tinh AB. AC l o i siiy ra cosẢ
2. Goi G la Irong lam ciia lam giiic ABC. Tinh AG. BC.
DAP AN THAM KHAO
Cau Dap an Diem
1
(1,0 diem) diem)
1. (0,5 diem) Xac djnh m de ham sọ..
1
(1,0
diem) Ham so xac dinh khi m" - 4x~ > 0 o
Vay tap xac dinh cua ham so lii D = (adsbygoogle = window.adsbygoogle || []).push({});
m X < 2 m m 2 ' 2 0,25 f,7
Ham so da cho xac dinh Iron (-2; 1] <=> (-2; 1] c D
m m < 2 va 1 < > 4 > 2 m >4<=> m < - 4 m > 4 0,25
2. (0,5 diem) Tim tap xac djnh cua ham so"...
2\\l\1 , Ham so y = — x a c dmh khi < Ham so y = — x a c dmh khi < y x- 3 x- 1 > 0 X- 3 ; ^ 0 0,25 x> l x^ 3 ' 0,25 II (2,0 diem)
V a y lap xac djnh ciia ham so lii D = |1; 3 )u( 3 ; + co).
1. (1,0 diem) Xac djnh a de ba dtf(/ng thang...
d| va d^cat nhau tai diem A ^ - a - 1 1 - 5 a - 1 5 ^
8 0,25 d|, d-,. d-, đng qui o d j d i qua A + 2a + 3 -5a -15 <=> = a ^ - a - 1 ' ^ 8 0,25 o - 5 a - 1 5 = -a- - 1 la + 16a + 2 4 o a - - 1 0 a - 3 9 = 0 <=> a = -3 hoac a = 13. Thírlai: . V d i a - - 3 : Ta Iha'yd2=d3.
• Vdi a = 13: Ta thay d,, d j , <ij la ba diTdng lhang phan biet va dong quị Vay a = 13 thoa miin dẹ
0,25
0,25
2. (1,0 diem) Xac dinh ham s d
Tap xiic dinh D = R. 0,25
Parabol y = 2x" + a x - b c 6 d i n h I ( - l ; 4) ncn: X = — 4 0,5 4 = 2 ( - l ) ^ + a ( - l ) - b a = 4 <=> < a = 4 a - 4 4 = 2 - a - b [4 = 2 - 4 - b [b = -6 Vay hiim so can lim y = 2x'^ + 4x + 6.
0,25
I l l
(2,0 diem) diem)
1. (0,5 diem) Chrfiij; niinh...
Tri{(fnn lufp Ị
+ N c i i a = 0. tCr giii thict siiy ra 2b + 3c = 0(1)
• NcLi h = 0, iir (1) => c = 0. K h i do phiTdng i r i n h da cho c6 n g h i c n i m o i x.
• Ncu b ^ 0, thi phifdng Irinh da cho c6 n g h i c m x = - — = - • b 3
TrU(!ni> hap 2.
T\i gia i h i c l siiy la h = - ^ • L i i c do
A = b- - 4ac = 5a + 3c •4ac = - ( 2 5 a - + 9 c - +4ac) 4
Z a. 2 c 2 2
+ 5c + — a 4 >().
TCr do siiy ra phifttng Irinh da cho c6 nghicm.
2. (0,75 diem) Giiii phifcTn^ t r i n h . . . Ta CO X " + - I ? = = 2 x ( x - 3 ) Ta CO X " + - I ? = = 2 x ( x - 3 ) X -+ x - 1 2 = 2 x ( x - 3 ) ncu x > 0 (1) X -- x - 1 2 = 2 x ( x - 3 ) ncu x < 0 (2) X > ( ) X - - 7 x + 12 = () x = 3 X = 4 X < ( ) x" - 5x + 12 = 0 (v6 nghicm). (adsbygoogle = window.adsbygoogle || []).push({});
Vay phiTcJog trinh da cho c6 hai nghiC-m x = 3, x = 4 •
3. (0,75 diem) G i a i phifi/n^ trinh . . .
Ta CO ( x " - 3 x ) \ / 4 - x - 3x" = 0 o 4 - x - 3 x " = ( ) (1) 4- X- 3 X - > ( ) x^ - 3 x = 0 (2) (!) o X = 1 hoac x = - - . 64 (2)C:> 4 - x - 3 x - > 0 X - - 3 x = 0 <=> i 4 - x - 3 x - > 0 x = 0 hoSc X = 3
Vay phirong trinh da cho c6 ba nghicm x = - —, x = 0, x = 1.
0,25
1. (1,0 diem) Chtfng niinh ttf giac A B C D la hinh c h g nhat.
T a c o A B = (1; 2). D C = (1; 2), BC = (4; - 2 ) . 0,5
Ta thiíy
chuTnhat.
A B = D C - ( I ; 2) A B . B C = 1.4 + 2 . ( - 2 ) = 0
n c n It? giac A B C D la hinh
2. (1,0 diem) Ílnh c h u vi, dien tich hinh chff nhat t r e n .
Ta CO A B = (1; 2) => A B = -JV+A = sjs.
B C = ( 4 ; - 2) ^ B C = 7i6 + 4 = ^/20 =275.
Chu v i cua hinh chff nhat A B C D :
C V A B C U - 2 ( A B + B C ) = 2.3v/5 - (dvcd).
D i e n tich ciia hinh chỉ nhat A B C D :
S ^ B c o = A B . B C = 10(dvdl)
1. (1,0 diem) G i a i hg phtfcfng trinh...
N h a n hai vc ciia phifcJng i r i n h (1) v d i 3 r o i cong v d i phffcfng trinh (2) i h c o v c , nhan hai véciia phiMng Irinh (1) v d i 4 r o i cong v(^i phifttng trinh (3) theo v c t a di/tJc:
7 x - 8 y = 2 2 fx = - 6 - y = 2 y = - 8 (5) The (5) viio (1) ta di/dc /. = - ! .
Vay he phiTiJng trinh da cho cd nghicm ( - 6; - 8; - 7).
2. (1,0 diem) Tim gia trj l<tn nhSt..
A p diing B D T Co-si cho 2 so khong a m 9, x - 9 , ta c6:
3 3 2 6 r^ t ' V x ^ ^ 1